Question: Find $\lim_{x\to -4 }\dfrac{x^2-16}{x^2+4x}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $-2$ (Choice C) C $0$ (Choice D) D The limit doesn't exist
Solution: Substituting $x=-4 $ into $\dfrac{x^2-16}{x^2+4x}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{x^2-16}{x^2+4x}$ can be simplified as $\dfrac{x-4}{x}$, for $x\neq -4 $. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-4 $. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x^2-16}{x^2+4x}=\dfrac{x-4}{x}$ for all $x$ -values in the interval $(-5,-3)$ except for $x=-4 $. Therefore, $\lim_{x\to -4 }\dfrac{x^2-16}{x^2+4x}=\lim_{x\to -4 }\dfrac{x-4}{x}=2$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -4 }\dfrac{x^2-16}{x^2+4x}=2$.